Difference between revisions of "2016 AMC 12A Problems/Problem 23"

Revision as of 13:42, 5 February 2016

Problem

Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?

$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$

Solution

Solution 1: Logic

WLOG the largest number is 1. Then the probability that the other two add up to at least 1 is $1/2$ .

Thus the answer is $1/2$ .

Solution 2: Calculus

When $a>b$ , consider two cases:

1) $0<a<\frac{1}{2}$ , then $\int_{0}^{\frac{1}{2}} \int_{0}^{a}2b \,\text{d}b\,\text{d}a=\frac{1}{24}$

2) $\frac{1}{2}<a<1$ , then $\int_{\frac{1}{2}}^{1} \left(\int_{0}^{1-a}2b \,\text{d}b + \int_{1-a}^{a}1+b-a \,db\right)\text{d}a=\frac{5}{24}$

$a<b$ is the same. Thus the answer is $\frac{1}{2}$ .

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
 ==Problem==
-Three numbers in the interval [0,1] are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
+Three numbers in the interval <math>\left[0,1\right]</math> are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
 <math>\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56</math>

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